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The distance of the point (1,0,2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

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Correct option is D

x−23=y+14=z−212=tP(3t+2,4t−1,12t+2) lies on x−y+z=16⇒t=1P(5,3,14)A(1,0,2)

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The distance of the point (1,0,2) from the point of intersection of the line x−23=y+14=z−212 and the plane x−y+z=16 is

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The distance of the point (1,0,2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is